Question: Multiply the following complex numbers: $({2+3i}) \cdot ({-2+5i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2+3i}) \cdot ({-2+5i}) = $ $ ({2} \cdot {-2}) + ({2} \cdot {5}i) + ({3}i \cdot {-2}) + ({3}i \cdot {5}i) $ Then simplify the terms: $ (-4) + (10i) + (-6i) + (15 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (10 - 6)i + 15i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (10 - 6)i - 15 $ The result is simplified: $ (-4 - 15) + (4i) = -19+4i $